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Interval of convergence

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Say that you were using ration test for
    \sum_{n=1}^\infty\frac{(-1)^{n+1} (x-4)^n}{n9^n!}\

    2. Relevant equations

    3. The attempt at a solution
    You take the limit of the above you will get

    ##\frac {1}{9} |x-4|##

    Book says radius of convergence is 9? Is this done by using the ratio test and making an inequality?

    ##\frac {1}{9} |x-4| <1##
    ##|x-4| < 9##

    OK then the book says in the next line the

    ##-5< x< 13##
    How did they get this? I thought that they maybe said if R = 9 then put ,-9 and 9 in the inequality. So,
    ##|(-9)-4| = -13## and ## |(9)-4| = 5##
    But that doesn't work because it is backwards theirs was -5 and 13 not 13 and -5
    What is going on. If you can elaborate the idea of radius of convergence and how they determined the interval please!
  2. jcsd
  3. Jul 14, 2013 #2
    HA! nevermind
  4. Jul 15, 2013 #3


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    For others who might be wondering, yes, this is correct. The "ratio test" says that a series , [itex]\sum a_n[/itex] converges (absolutely) as long as
    [tex]\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|< 1[/tex]

    Here, that ratio is [itex][|x- 4|^{n+1}]/[(n+1)9^{n+1}][n 9^n]/[|x- 4|^n= [n/(n+1)]|x- 4|/9[/itex]. (I have ignored the "[itex](-1)^{n+1}[/itex]" because of the absolute value.)

    The limit as n goes to infinity is, as Jbreezy said, |x- 4|/9. From |x- 4|/9< 1 we obviously get |x- 4|< 9 and so -9< x- 4< 9. The radius of convergence, the distance from the center point, x= 4. to each end of the interval of convergence is 9. So, adding 4 to each part,-9+ 4= -5< x< 13= 9+ 4.
    Your "error" is trying to incorporate the absolute value where it does not belong.

    Now, the series might or might (or might converge but not "absolutely") at x= -5 and x= 13. Those you would have to check separately.

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